Question

The minimum number of times one has to toss fair coins so that the probability of observing at least one head is at least 90% is:
(A) 5
(B) 3
(C) 2
(D) 4

Answer 1

Assume the number of tosses of the coin is n. The probability of getting a head and getting a tail is $\dfrac{1}{2}$ . Here, we have only two possible outcomes, either head or tail. It means binomial distribution can be applied here. We know the formula for the binomial distribution, the probability of getting x number of head out of n tosses = $^{n}{{C}_{x}}{{\left( \dfrac{1}{2} \right)}^{x}}{{\left( \dfrac{1}{2} \right)}^{n-x}}$ . We know that in an event the sum of the probabilities of all possible outcomes is 1. So, $P\left( H=0 \right)+P\left( H=1 \right)+P\left( H=2 \right)+P\left( H=3 \right)+..............+P\left( H=n \right)=1$ . Now, transform the equation as $P\left( H=0 \right)+P\left( H\ge 1 \right)=1$ . Now, using the binomial distribution formula, get the value of $P\left( H=0 \right)$ . Then, put the value of $P\left( H=0 \right)$ in the equation $P\left( H=0 \right)+P\left( H\ge 1 \right)=1$ and get the value of $P\left( H\ge 1 \right)$ . It is given that the probability of at least one head is 90%. It means we have $P\left( H\ge 1 \right)>0.9$ . Now, solve it further and get the value of n.

Complete step by step answer:
First of all, let us assume the number of times that the man tosses the coin is n.
In a coin, we have only two possible outcomes, one is head and the other is tail.
The probability of getting a head in a fair coin = $\dfrac{1}{2}$ ……………………………….(1)
The probability of getting a tail in a fair coin = $\dfrac{1}{2}$ ……………………………….(2)
Now, according to the question, we have the probability of getting at least one head is at least 90%.
It means the probability of getting at least one head can be greater than 90%.
$\text{The probability of getting at least one head }\ge \text{ }0.9$ ……………………………….(3)
For n tosses of the coin, we have to get at least one head.
We have a coin in which there are only two outcomes, either head or tail.
It means here we have a binomial distribution.
We know the formula for the binomial distribution, the probability of getting x number of head out of n tosses = $^{n}{{C}_{x}}{{\left( \dfrac{1}{2} \right)}^{x}}{{\left( \dfrac{1}{2} \right)}^{n-x}}$ ……………………………..(4)
We know that in an event the sum of the probabilities of all possible outcomes is 1.
For n tosses of a coin, we can have 0 head, 1 head, 2 heads, ……………………., up to n heads.
So, we can say that the summation of the probability of getting exactly 0 head that is, $P\left( H=0 \right)$ , probability of getting exactly 1 head that is, $P\left( H=1 \right)$ , probability of getting exactly 2 heads $P\left( H=2 \right)$ , ……………………., probability of getting exactly n heads $P\left( H=n \right)$ is 1.
In the mathematical form, we can write it as,
$P\left( H=0 \right)+P\left( H=1 \right)+P\left( H=2 \right)+P\left( H=3 \right)+..............+P\left( H=n \right)=1$ …………………………….(5)
Transforming equation (5), we get
$P\left( H=0 \right)+P\left( H\ge 1 \right)=1$
$\Rightarrow P\left( H\ge 1 \right)=1-P\left( H=0 \right)$ ……………………………..(6)
In equation (6), $P\left( H\ge 1 \right)$ is the probability of getting at least one head and $P\left( H=0 \right)$ is the probability of getting exactly zero head.
Using the formula that is shown in equation (4), we can get the probability of exactly zero heads,
The probability of getting exactly zero head = $^{n}{{C}_{0}}{{\left( \dfrac{1}{2} \right)}^{0}}{{\left( \dfrac{1}{2} \right)}^{n-0}}=1\times 1\times {{\left( \dfrac{1}{2} \right)}^{n}}={{\left( \dfrac{1}{2} \right)}^{n}}$ ………………………….(7)
From equation (6) and equation (7), we get
$\Rightarrow P\left( H\ge 1 \right)=1-{{\left( \dfrac{1}{2} \right)}^{n}}$ ……………………………(8)
From equation (3) and equation (8), we have the probability of getting at least one head.
Now, comparing equation (3) and equation (8), we get

& \Rightarrow 1-{{\left( \dfrac{1}{2} \right)}^{n}}\ge 0.9 \\\ & \Rightarrow 1-0.9\ge {{\left( \dfrac{1}{2} \right)}^{n}} \\\ & \Rightarrow 0.1\ge {{\left( \dfrac{1}{2} \right)}^{n}} \\\ & \Rightarrow \dfrac{1}{10}\ge \dfrac{1}{{{2}^{n}}} \\\ \end{aligned}$$ $$\Rightarrow {{2}^{n}}\ge 10$$ ……………………(9) We have to take the value of n in such a way that $${{2}^{n}}>10$$ . We know that $${{2}^{3}}=8$$ and $${{2}^{4}}=16$$ ……………………….(10) Now, from equation (9) and equation (10), we can figure out that the value of n should be greater than 3. So, the value of n is greater than equal to 4, $$n\ge 4$$ ……………………(11) From equation (10), we can get the minimum value of n. Therefore, the minimum value of n is 4. **So, the correct answer is “Option D”.** **Note:** In this question, one can think to find the value of the probability of getting at least one head by the summation of the probability of getting exactly one head, the probability of getting exactly two heads, …………………, and the probability of getting exactly n heads. If we do so, then our calculation will become too complex. So, don’t approach this question by this method.